0
我用現有的行在我的表user_table中添加了一個新的uuid字段。由於競爭條件,我在遷移時收到「重複輸入」錯誤。然後我遵循的步驟在這裏:Django:在uuid4中重複輸入
http://django.readthedocs.org/en/latest/howto/writing-migrations.html
現在我的第一和第二遷移運行,但我最後的遷徙給了我同樣的錯誤。是我的每個鏈接3遷移如下:
遷移 '0009_label'
from __future__ import unicode_literals
from django.db import models, migrations
import uuid
class Migration(migrations.Migration):
dependencies = [
('app', '0008_label'),
]
operations = [
migrations.AddField(
model_name='user_table',
name='UUID_loc',
field=models.UUIDField(default=uuid.uuid4, null=True),
),
migrations.AlterField(
model_name='another_table',
name='Time',
field=models.CharField(default=0, max_length=3),
),
]
遷移 '0010_label'
from __future__ import unicode_literals
from django.db import migrations, models
import uuid
def gen_uuid(apps, schema_editor):
MyModel = apps.get_model('app', 'user_table')
for row in MyModel.objects.all():
row.uuid = uuid.uuid4()
row.save()
class Migration(migrations.Migration):
dependencies = [
('app', '0009_label'),
]
operations = [
# omit reverse_code=... if you don't want the migration to be reversible.
migrations.RunPython(gen_uuid, reverse_code=migrations.RunPython.noop),
]
遷移 '0011_label'
from __future__ import unicode_literals
from django.db import models, migrations
import uuid
class Migration(migrations.Migration):
dependencies = [
('app', '0010_label'),
]
operations = [
migrations.AlterField(
model_name='user_table',
name='UUID_loc',
field=models.UUIDField(default=uuid.uuid4, unique=True),
),
]
的聯繫是與我非常相關,但不幸的是我得到了同樣的錯誤。現在我陷入了這裏,我的表中有UUID_loc字段,但它還不是唯一的,即第三個遷移尚未運行。任何人都可以提供一些見解嗎?謝謝。
謝謝你的回答,但我仍然得到相同的錯誤:/ – plumSemPy
更新:我意識到我有一個列名稱的差異,它現在必須工作:) row.uuid = uuid.uuid4()應該是行。 UUID_loc = uuid.uuid4();那對我來說很愚蠢 – plumSemPy